A liquid of mass M and specific heat S is at a temperature 2 t. If another liquid of mass M and thermal capacity 1.5 times, at a temperature of t/3 is added to it, the resultant temperature will be:

Let the common temperature of the mixture be T.For first liquid, mass, M, specific heat of liquid, S and initial temperature of liquid, 2t.For second liquid, mass, M, specific heat of liquid, 1.5 S and initial temperature of liquid, t3Heat loss by the hot liquid = Heat consumed by cold liquidMS(2t−T)=M(1.5S)T−t3⇒2t−T=1.5T−t2⇒52T=52t⇒T=t